How to calculate this integral when the function is unknown? Let $f :[0,1]\to \mathbb{R}$ with $f'(x) = \sqrt{1+f^2(x)}$ for all $x \in [0,1]$. If $f(0) + f(1) = 0,$ calculate the integral
$$I=\int_{0}^{1}f(x)dx$$
Any help would be appreciated. Thanks.
 A: Step I) $f$ is a real valued function, thus, for all $x \in [0,1]$,
$$
1=\frac{f'(x)}{\sqrt{1+f(x)^2}}
$$
Step II)
Evaluate the definite integral
$$
\int_0^1 f(x)\,dx=\int_0^1 \frac{f(x)f'(x)}{\sqrt{1+f(x)^2}}\,dx
$$
Step III) Note that the integrand is the derivative of $\sqrt{1+f(x)^2}$
A: You could do it by solving the separable differential equation $y' = \sqrt{1+y^2}$, but you could also use symmetry: show that $f(1/2 + t) + f(1/2-t) = 0$.
EDIT: Oh, the easiest way:  $$ \dfrac{d}{dx} \sqrt{1+f(x)^2} = \ldots$$ 
A: From the differential equation $f' = \sqrt{1 + f^2}$, we obtain the differential equation $(f')^2 - f^2 = 1$. Since $\cosh^2 (x) - \sinh^2 (x) = 1$, we look for solutions of the following form
$$f (x) = c_1 \,\mathrm{e}^{x} + c_2 \, \mathrm{e}^{-x}$$
Using $(f')^2 - f^2 = 1$, we obtain $c_1 c_2 = -\frac{1}{4}$. From $f (0) + f (1) = 0$, we obtain $c_2 = - \mathrm{e} \, c_1$.
Thus,
$$f (x) = \pm\sinh \left(x - \frac{1}{2}\right)$$
which is odd-symmetric about $x = \frac{1}{2}$. Hence, $\displaystyle\int_0^1 f (x) \, \mathrm{d}x = 0$. 
